(1-5i)(-9+2i)=0

Solution for (1-5i)(-9+2i)=0 equation:

(1-5i)(-9+2i)=0

We add all the numbers together, and all the variables

(-5i+1)(2i-9)=0

We multiply parentheses ..

(-10i^2+45i+2i-9)=0

We get rid of parentheses

-10i^2+45i+2i-9=0

We add all the numbers together, and all the variables

-10i^2+47i-9=0

a = -10; b = 47; c = -9;
Δ = b2-4ac
Δ = 472-4·(-10)·(-9)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-43}{2*-10}=\frac{-90}{-20}
=4+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+43}{2*-10}=\frac{-4}{-20}
=1/5
$